Cho phương trình: 3x^2 - (m + 3)x + 2 = 0 a) Giải phương trình khi m = 2

a) Khi m = 2, phương trình thành: $3x^2-5x+2=0$

Vì $a+b+c=3-5+2=0\Rightarrow \left\{\begin{array}{l}{x}_1=1\\ x_2=\frac{2}{3}\end{array}\right.$

b) $3x^2-\left(m+3\right)x+2=0$

$\Delta ={\left(m+3\right)}^2-\mathrm{4.3.2}=m^2+6m-15$

Để phương trình có nghiệm $\iff \Delta \ge 0\iff m^2+6m-15\ge 0\iff \left[\begin{array}{l}m\le -3-2\sqrt{6}\\ m\ge -3+2\sqrt{6}\end{array}\right.$

Khi đó, theo hệ thức Viet ta có: $\left\{\begin{array}{l}{x}_1+x_2=\frac{m+3}{3}\\ x_1{x}_2=\frac{2}{3}\end{array}\right.$. Ta có:

$\begin{array}{l}{x}_1+x_1{x}_2+x_2=4\\ \Rightarrow \left(x_1+x_2\right)+x_1{x}_2=4\\ \iff \frac{m+3}{3}+\frac{2}{3}=4\Rightarrow m=7\left(tm\right)\end{array}$

$\begin{array}{l}c)B=x_1^2+x_2^2-6x_1-6x_2\\ ={\left(x_1+x_2\right)}^2-2x_1{x}_2-6\left(x_1+x_2\right)\\ ={\left(\frac{m+3}{3}\right)}^2-2.\frac{2}{3}-6.\frac{m+3}{3}\\ =\frac{m^2+6m+9}{9}-\frac{4}{3}-2\left(m+3\right)\\ =\frac{m^2}{9}-\frac{4}{3}m-\frac{19}{3}\end{array}$

$\begin{array}{l}={\left(\frac{m}{3}\right)}^2-2.\frac{m}{3}.2+4-\frac{31}{3}\\ ={\left(\frac{m}{3}-2\right)}^2-\frac{31}{3}\ge -\frac{31}{3}\end{array}$

$\Rightarrow MinB=\frac{-31}{3}\iff m=6$